integration of a log-normal random variable Let X be a log-normal disributed random variable, $\forall{a}\in\mathbb{R}$ compute: $$\int_{\mathbb{R}}|a+X|dP.$$
I get that when $a\ge0$, since $X$'s support is $[0,\infty)$, holds:
$$\int_{\mathbb{R}}|a+X|dP=\int_{\mathbb{R}}(a+X)dP=a+E[X].$$
When $a<0$, is there a way of computing the integral without using the pdf like I did before?
 A: Note $|a+X|=(a+X)^+-(a+X)^-=\max(a+X,0)-\min(a+X,0)$. If $a<0$ set $b:=-a,\,b>0$ and we have
$$|X-b|=(X-b)^+-(X-b)^-=\max(X-b,0)-\min(X-b,0)$$
The expectation $E[(X-b)^+]$ is given in closed form by the Black-Scholes formula for call options. On the other hand:
$$E[(X-b)^-]=-E[(b-X)^+]$$
which is again given by the Black-Scholes formula of put options.
A: If $a < 0$, let $b = -a$, so that $|a + X| = |X - b|$ and $b > 0$.  Then clearly, when $X \ge b$, $|X - b| = X - b$, and when $0 \le X < b$, we have $|X - b| = -(X-b) = b-X$. To summarize,
$$|X-b| = \begin{cases} b-X, & 0 \le X < b \\ X-b, & X \ge b. \end{cases}$$
The structure of this function suggests the limited expected value; i.e. recall $X \wedge b = \min(X, b)$, and
$$\begin{align}
\operatorname{E}[X \wedge b] 
&= \int_{x=0}^\infty \min(X, b) f_X(x) \, dx \\
&= \int_{x=0}^b x f_X(x) \, dx + b \int_{x=b}^\infty f_X(x) \, dx \\
&= \int_{x=0}^b x f_X(x) \, dx + b S_X(b) \tag{1}
\end{align}$$ where $S_X(x)$ is the survival function of $X$.
Hence
$$\begin{align}
\int_{x = 0}^\infty |x-b| f_X(x) \, dx 
&= \int_{x=0}^b (b-x) f_X(x) \, dx + \int_{x=b}^\infty (x-b) f_X(x) \, dx \\
&= b F_X(b) - 2 \int_{x=0}^\infty x f_X(x) \, dx + \operatorname{E}[X] - b S_X(b) \\
&= b F_X(b) - 2 (\operatorname{E}[X \wedge b] - b S_X(b)) + \operatorname{E}[X] - b S_X(b) \\
&= b (F_X(b) + S_X(b)) - 2 \operatorname{E}[X \wedge b] + \operatorname{E}[X] \\
&= b - 2 \operatorname{E}[X \wedge b] + \operatorname{E}[X].  \tag{2}
\end{align}$$
Unfortunately, there isn't a more convenient form; all we have really done here is express $\operatorname{E}[|X - b|]$ in terms of the expectation of a different function, namely $X \wedge b$.
